How angles transform under Lorentz transforms - a thought experiment Lets say I have two frames, $S$ and $S'$; they are in standard configuration and $S'$ is moving with speed $v$ away from $S$. I have a pipe in $S'$, which is at angle $\theta'$ from its $x$ axis. I can calculate the angle which would be measured between the pipe and the $x$ axis in $S$: $\theta$. I note that $\theta \neq \theta'$.
However, if I shoot a photon at an angle $\theta'$ in $S'$, since the velocity of light is constant in all frames, this angle will be the same in $S$ as well. So in $S'$ the photon does not hit the wall of the pipe, but in $S$ it does, because it is not going at an angle $\theta'$, and the pipe is at angle $\theta$, and $\theta' \neq \theta$. Whether the photon hits the wall of the pipe or not should not depend on the frame, so I have made a mistake above.
What I think my mistake is: speed of light is constant, it does not mean that the components of the velocity in $x$ and $y$ directions is constant in all frames. Therefore if I shoot a photon at a certain angle in one frame, the angle won't necessarily be the same in another frame.
Since I would like to find that hitting the wall is Lorentz-invariant, shooting anything at any speed (ie not just photons) at an angle $\theta'$ in $S'$ is being seen as shot at an angle $\theta$ in $S$.
Is this reasoning correct? if not, which step in the reasoning is wrong? (I think it is not correct, but I haven't found the mistake yet.)
 A: The speed of light, not the velocity of light, is Lorentz-invariant.
Wikipedia discusses the relativistic aberration of light.
A: You're right that the (spatial) angle of the light beam in $S$ is not $θ'$. But it isn't $θ$ either.
If you look at constant-coordinate-time slices of $S$, the pipe is moving to the right (while pointing diagonally), while the light is moving diagonally. If the light moved diagonally at the same angle as the pipe is pointing, the pipe would "leave it behind" and it would hit the wall. The angle of the light's motion has to be closer to the horizontal for it to avoid hitting the pipe. If you work out the necessary angle (keeping in mind that the speed must be $c$), you'll get the relativistic aberration formula.
A: When dealing with the direction light travels, it's best to use the wave vector ${\bf k}$, and it's 4-vector form as the 4-gradient of the Lorentz scalar phase:
$$ k^{\mu} \equiv \partial^{\mu}\phi = (\frac 1 c\frac{\partial\phi}{\partial t}, {\bf \nabla}\phi)$$
For a plane wave:
$$ A(x,t) = Ae^{i\phi(x, t)}=Ae^{ik^{\mu}x_{\mu}}=Ae^{i({\bf k \cdot x} -\omega t)}$$
is:
$$k^{\mu} =(\omega/c, {\bf k}) = (\omega/c, k\sin{\theta},0,k\cos{\theta}) = k(1, \sin{\theta}, 0, \cos{\theta})$$
Here $\theta$ is the angle in unprimed frame.
A boost along $-z$ at $v$ is a straightforward Lorentz transformation:
$$\omega' = \gamma(\omega+\frac v c k\cos{\theta})$$
$$k_x' = k_x = k\sin{\theta}$$
$$k_y'=0$$
$$k_z'= k_z + \frac v {c^2}\omega = k\gamma(\cos{\theta}+\beta)$$
The primed angle is then:
$$ \tan{\theta'} = \frac{\sin{\theta}}{\beta + \cos{\theta}} $$
which is the standard result for stellar aberrations. (Note: you also get the relativistic Doppler shift from $\omega'/\omega$).
